Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry - Exercises - Page 183: 58


These reactants are capable of producing $2.33$ g of $BaSO_4$;

Work Step by Step

1. Write and balance the reaction between $BaCl_2$ and $Fe_2(SO_4)_3$: $BaCl_2(aq) + Fe_2(SO_4)_3(aq) -- \gt BaSO_4(s) +FeCl_3(aq)$ - Balance it: $3BaCl_2(aq) + Fe_2(SO_4)_3(aq) -- \gt 3BaSO_4(s) +2FeCl_3(aq)$ 2. Find the number of moles of $BaSO_4$ that each compound can produce. $100.0mL \times \frac{1L}{1000mL} \times \frac{0.100mol(BaCl_2)}{1L} \times \frac{3mol(BaSO_4)}{3mol(BaCl_2)} = 0.0100mol(BaSO_4)$ $100.0mL \times \frac{1L}{1000mL} \times \frac{0.100mol(Fe_2(SO_4)_3)}{1L} \times \frac{3mol(BaSO_4)}{1mol(Fe_2(SO_4)_3)} =0.0300mol(BaSO_4)$ - $BaCl_2$ produces less barium sulfate, so, it is the limiting reactant. - Use the number of moles produced by $BaCl_2$ to calculate the mass: Molar mass: 137.3* 1 + 32.07* 1 + 16.00* 4 = 233.4g/mol $(BaSO_4)$ $0.0100 mol(BaSO_4) \times \frac{233.4g (BaSO_4)}{1mol(BaSO_4)} = 2.33g(BaSO_4)$
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